integral of 1+sin^2(x)

Lösung

\int 1 + sin^{2} (x) d x

\frac{3}{2} x - \frac{1}{4} sin (2 x) + C

Schritte zur Lösung

\int 1 + sin^{2} (x) d x

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 1 d x + \int sin^{2} (x) d x

\int 1 d x = x

\int sin^{2} (x) d x = \frac{1}{2} (x - \frac{1}{2} sin (2 x))

= x + \frac{1}{2} (x - \frac{1}{2} sin (2 x))

Vereinfache

x + \frac{1}{2} (x - \frac{1}{2} sin (2 x)) : \frac{3}{2} x - \frac{1}{4} sin (2 x)

= \frac{3}{2} x - \frac{1}{4} sin (2 x)

Füge eine Konstante zur Lösung hinzu

= \frac{3}{2} x - \frac{1}{4} sin (2 x) + C

\int \frac{1}{9 - x} d x

\int sin (\frac{3 x}{2}) d x

\int sin^{7} (x) cos^{3} (x) d x

x \to 0 lim (x^{19 x})

\int_{1}^{\infty} f (x) d x